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Question: In \(1L\)of a \(3M\)solution of sodium chloride, how many grams of sodium chloride are there in the ...

In 1L1Lof a 3M3Msolution of sodium chloride, how many grams of sodium chloride are there in the solution?

Explanation

Solution

We all know that molarity is defined as the amount of moles of a compound dissolved in an amount of solvent. It is expressed in units mol L−1mol\,{L^{ - 1}}. If we are given with volume and amount values, it is quite clear that we have to use the molarity concept. We should be quite clear about all the concentration terms.

Formula used: M=nVM = \dfrac{n}{V}

Complete step-by-step answer:
We know,
M=nVM = \dfrac{n}{V}
Where n is the number of moles of solute or amount of the solute in moles.
V is the volume of the solution in litres.
M is the molarity of the solution in units of mol L−1mol\,{L^{ - 1}}
Substituting the values given in the question in the formula,
M=3M V=1L  M = 3M \\\ V = 1L \\\
n=M×V =3M×1L=3molL−1×1L=3mol  n = M \times V \\\ = 3M \times 1L = 3mol{L^{ - 1}} \times 1L = 3mol \\\
No. of moles of sodium chloride is equal to three.
But we have to find the grams of sodium chloride present in the solution. For this, we need to convert the number of moles into weight by using a relation with the molecular mass of the substance.
Molecular mass of sodium Na=23g mol−1 = 23g\,mo{l^{ - 1}}
Molecular mass of chloride Cl=35.5 g mol−1 = 35.5\,g\,mo{l^{ - 1}}
Molecular mass of sodium chloride NaCl=23+35.5=58.5 g mol−1 = 23 + 35.5 = 58.5\,g\,mo{l^{ - 1}}
Using,
n=wMn = \dfrac{w}{M}
M=58.5g mol−158.5g\,mo{l^{ - 1}}
n=3  = 3\,mol
w=n×M=3 mol×58.5 g mol−1 =175.5gw = n \times M = 3\,mol \times 58.5\,g\,mo{l^{ - 1}}\, = 175.5g
Hence, 175.5175.5 g is the required weight of sodium chloride in solution.

Note: You should know the difference between molarity, molality, normality and all other concentration terms as sometimes these values can also be given. Standard units are used in the question but sometimes we have to do unit conversions to arrive at the answer.
You should know which concentration terms are temperature dependent and which are temperature independent. Since molarity is volume dependent which is further temperature dependent which makes molarity also temperature dependent while molality is temperature independent as it does not contain any temperature dependent quantity.